A) 8.0m
B) 5.5m
C) 6.0 m
D) 6.5 m
Correct Answer: B
Solution :
From the lens formula (for first lens) \[\frac{1}{{{f}_{1}}}=\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}\] \[\Rightarrow \] \[\frac{1}{2}=\frac{1}{{{v}_{1}}}-\frac{1}{(-4)}\] \[\Rightarrow \] \[\frac{1}{2}+\frac{1}{4}=\frac{1}{{{v}_{1}}}=\frac{3}{4}\] ?(i) \[\Rightarrow \] \[{{v}_{1}}=\frac{4}{3},\,{{u}_{2}}=3-\frac{4}{3}=5/3\] This image will be treated as the source for second lens, then again from lens formula, we have \[\frac{1}{{{f}_{2}}}=\frac{1}{{{v}_{2}}}-\frac{1}{{{u}_{2}}}\] \[\Rightarrow \] \[\frac{1}{1}=\frac{1}{{{v}_{2}}}+5/3\] [By Eq. (i)] \[\Rightarrow \] \[1-5/3=\frac{1}{{{v}^{2}}}\Rightarrow -3/2\] This is the final image distance from 2nd lens. So, the overall distance of image from the primary source(or object) Let \[d=4+3-1.5=5.5\]
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